Factor completely. $(3x^2-12x)(x^2-2x+1)=$
The expression is already somewhat factored, because it is given as the product of two factors, $(3x^2-12x)$ and $(x^2-2x+1)$. To completely factor the expression, we need to factor each of these factors further. Factoring $(3x^2-12x)$ These terms have a common factor. The greatest common factor of $3x^2$ and $-12x$ is $3x$. Let's factor $3x$ out of $3x^2-12x$ : $\begin{aligned} &\phantom{=}3x^2-12x \\\\ &=3x(x)+3x(-4) \\\\ &=3x(x-4) \end{aligned}$ Factoring $(x^2-2x+1)$ We notice this expression has the perfect square pattern: $\begin{aligned} &\phantom{=}x^2-2x+1 \\\\ &=(x)^2+2(x)(-1)+(-1)^2 \\\\ &=(x-1)^2 \end{aligned}$ [Is there another way to factor this?] Putting it all together $\begin{aligned} &\phantom{=}{(3x^2-12x)}C{(x^2-2x+1)} \\\\ &={3x(x-4)}C{(x-1)^2} \end{aligned}$ In conclusion, this is the completely factored expression: $3x(x-4)(x-1)^2$